### Video Transcript

In this video, weβll learn how to
find the derivative of the natural logarithmic function. Weβll begin by recalling the
properties of logarithms that weβre going to use in this video before looking at how
these properties can help us to find the derivative of simple logarithmic
functions. Weβll then consider how we can
apply these processes in conjunction with the standard rules for differentiation to
find the derivative of more complicated logarithmic functions. Letβs begin by recalling one of the
key properties of logarithms that weβre going to be using throughout this video. That is, log base π of π₯ to the
power of π¦ is the same as π¦ times log base π of π₯.

Since the natural logarithm is just
log base π, then we can say that the natural logarithm of π₯ to the power of π¦ is
the same as π¦ times the natural logarithm of π₯. Weβll also lean heavily on the use
of the chain rule, product rule, and quotient rule. The chain rule says that if π¦ is a
function in π’ and π’ itself is a function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by
dπ’ times dπ’ by dπ₯. The product rule says that, for two
differentiable functions π’ and π£, the derivative of their product π’π£ is equal to
π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. And finally, we recall the quotient
rule. This says the derivative of the
quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus π’
times dπ£ by dπ₯ all over π£ squared.

If π¦ is equal to the natural log
of π₯, what is dπ¦ by dπ₯?

There are two ways we can find the
derivative of the natural log of π₯. The first is using differentiation
from first principles. Alternatively, we can recall the
following facts about functions that are inverses of one another. If π of π₯ and π of π₯ are
inverses of each other, then we can say that π prime of π₯, the derivative of our
function π, is equal to one over π prime of π of π₯. This is really useful as we know
the exponential function π to the power of π₯ and the natural logarithmic function
are inverses of one another. We can also quote the general
result that the derivative of π to the power of π₯ is simply π to the power of
π₯. So letβs let π of π₯ be equal to
π to the power of π₯ and π of π₯ be equal to the natural log of π₯, then π prime
of π₯ is equal to one over π prime of π of π₯.

We saw that π prime of π₯ was
equal to π the power of π₯. So this means that π prime of π
of π₯ is equal to π to the power of π of π₯, which is equal to π to the power of
the natural log of π₯. Well, π to the power of the
natural log of π₯ is simply π₯. So we obtain π prime of π₯ to be
equal to one over π₯. This means the derivative of the
natural logarithm of π₯ is one over π₯ where π₯ is greater than zero. Now, this requirement on π₯ is
necessary for two reasons. Firstly, π₯ has to be greater than
zero for the natural logarithm of π₯. So the same must be true for its
derivative. But we also know that if π₯ is
equal to zero, then our derivative is one over zero which is undefined. We could actually redefine this and
say that the derivative of the natural logarithms of the absolute value of π₯ is one
over π₯ when π₯ is not equal to zero. Weβll be quoting this general
result throughout the remainder of this video.

Find dπ¦ by dπ₯, given that π¦ is
equal to the natural log of two π₯ plus seven cubed.

In this question, we have a
function of a function, in other words, a composite function. We can use the chain rule to find
the derivative of a composite function, though since weβre working with the natural
logarithm, itβs sensible to first consider whether there is anything we can do to
manipulate our expression before differentiating. We recall the general result for
the natural logarithm of a power. The natural logarithm of π₯ to the
π¦th power is π¦ times the natural logarithm of π₯. And we see that we can now rewrite
our equation as π¦ equals three times the natural logarithm of two π₯ plus
seven. This is great because we know that
the derivative of a constant multiple of an expression in π₯ is equal to the
multiple of the derivative of that expression. In other words, the derivative of
three times the natural log of two π₯ plus seven is equal to three times the
derivative of the natural log of two π₯ plus seven.

But how do we differentiate the
natural log of two π₯ plus seven? Weβre going to quote the general
result for the derivative of the natural logarithm and weβre going to use the chain
rule. The chain rule says that if π¦ is a
function in π’ and π’ itself is a function in π₯, then the derivative of π¦ with
respect to π₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. And we also know that the
derivative of the natural log of π₯ is simply one over π₯. So we let π’ be equal to two π₯
plus seven such that π¦ is equal to the natural log of π’. We know that dπ’ by dπ₯ is two and
dπ¦ by dπ’ is one over π’. The derivative then of the natural
log of two π₯ plus seven is the product of these. Itβs two times one over π’. But we replace π’ with two π₯ plus
seven.

And we see that the derivative of
the natural log of two π₯ plus seven is two over two π₯ plus seven. We recall we said that the
derivative of three times the natural log of two π₯ plus seven is three times the
derivative of the natural log of two π₯ plus seven. So the derivative is three times
two over two π₯ plus seven, which is simply six over two π₯ plus seven.

Now, there is actually a further
standard result that we can extract from this example. That is, if π¦ is equal to the
natural log of some function in π₯, then dπ¦ by dπ₯ is equal to π prime of π₯, the
derivative of that function, over the function itself, π of π₯. If we go back to our example, we
were trying to find the derivative of the natural log of two π₯ plus seven. We obtained that as being two over
two π₯ plus seven. Thatβs the derivative of two π₯
plus seven over two π₯ plus seven. Letβs consider another example.

If π of π₯ is equal to three times
the natural log of two π₯ plus four times the natural log of π₯, find π prime of
one.

Here, we have a composite
function. And that means we can use the chain
rule to find π prime of π₯, the derivative of the function. Alternatively, thereβs a general
result that we can quote. That is, if π of π₯ is the natural
log of some other function π of π₯, then the derivative π prime of π₯ is equal to
π prime of π₯ over π of π₯. Now, we can kind of forget about
the three for a moment. We know that the derivative of
three times the natural log of two π₯ plus four times the natural log of π₯ will be
equal to three times the derivative of that natural logarithm function. So letβs let π of π₯ be equal to
two π₯ plus four times the natural log of π₯. Then we know that the derivative π
prime of π₯ is equal to two plus four times one over π₯. And thatβs because the derivative
of the natural log of π₯ is simply one over π₯.

We can simplify this and we see
that π prime of π₯, the derivative of two π₯ plus four times the natural log of π₯
is two plus four over π₯. π prime of π₯, the derivative of
our function, is therefore three times π prime of π₯ over π of π₯. Thatβs three times two plus four
over π₯ over two π₯ plus four times the natural log of π₯. Now, usually weβd look to simplify
this expression somewhat. But here weβre being asked to find
the value of π prime of one. So we substitute π₯ equals one into
our expression for the derivative. And we get three times two plus
four over one over two times one plus four times the natural log of one. Well, we know the natural log of
one is zero. So four times the natural log of
one is also zero. And we see that π prime of one
becomes three times six over two which is nine. π prime of one is equal to
nine.

In our next examples, weβll
consider how we can use other rules for differentiation to evaluate the derivative
of natural logarithmic functions.

Find dπ¦ by dπ₯, given that π¦
equals four times the natural log of π₯ plus three over four times the natural log
of π₯ minus seven.

In this question, we have a
fraction or a quotient. This tells us we can use the
quotient rule to find the derivative dπ¦ by dπ₯. This says that the derivative of
the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus
π’ times dπ£ by dπ₯ all over π£ squared. We let π’ be equal to four times
the natural log of π₯ plus three. And π£ is equal to the denominator
of our fraction. Thatβs four times the natural log
of π₯ minus seven. We then quote the general result
for the derivative of the natural log of π₯; itβs one over π₯. And since the derivative of a
constant is zero, we see that dπ’ by dπ₯ is equal to four lots of one over π₯, which
is simply four over π₯. And similarly dπ£ by dπ₯ is also
four over π₯. dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯
all over π£ squared.

Letβs multiply the numerator and
denominator of this fraction by π₯ to simplify. When we do, we see that dπ¦ by dπ₯
is equal to four times four times the natural log of π₯ minus seven minus four times
four times the natural log of π₯ plus three all over π₯ times four times the natural
log of π₯ minus seven squared. We distribute the parentheses on
our numerator. And we see that we have 16 times
the natural log of π₯ minus 16 times the natural log of π₯ which gives us zero. And we found the derivative of our
quotient. Itβs negative 40 over π₯ times four
times the natural log of π₯ minus seven squared.

Find the first derivative of the
function π¦ equals negative seven π₯ to the fourth power times the natural log of
six π₯ to the fourth power.

Here, we have the product of two
differentiable functions. The first is negative seven π₯ to
the fourth power and the second is the natural log of six π₯ of the fourth
power. We can, therefore, find the first
derivative of our function by using the product rule. This says that the derivative of
the product of two differentiable functions, π’ and π£, is π’ times dπ£ by dπ₯ plus
π£ times dπ’ by dπ₯. If we let π’ be equal to negative
seven π₯ to the fourth power, then dπ’ by dπ₯ is equal to four times negative seven
π₯ cubed. Thatβs negative 28π₯ cubed. We then let π£ be equal to the
natural log of six π₯ to the fourth power.

So how do we obtain dπ£ by dπ₯? Well, we can do one of two
things. We could spot that we have a
composite function and use the chain rule to find its derivative. Alternatively, we can quote the
general result for the derivative of the logarithm of some function π of π₯. This says that if π¦ is equal to
the natural log of this function π of π₯, then dπ¦ by dπ₯ is equal to π prime of
π₯, the derivative of that function, divided by the original function π of π₯. In this case, π of π₯ is equal to
six π₯ to the fourth power. So π prime of π₯, the derivative
of this, is four times six π₯ cubed, which is 24π₯ cubed. dπ£ by dπ₯ is, therefore,
24π₯ cubed, divided by the original function, six π₯ to the fourth power.

Well, this simplifies quite nicely
to four over π₯. And we can now substitute
everything we have into the formula for the product rule. dπ¦ by dπ₯ is π’ times dπ£ by dπ₯. Thatβs negative seven π₯ to the
fourth power times four over π₯ plus π£ times dπ’ by dπ₯. Thatβs the natural log of six π₯ to
the fourth power times negative 28π₯ cubed. We simplify, and then we factor
negative 28π₯ cubed. And we obtained dπ¦ by dπ₯ to be
negative 28π₯ cubed times the natural logarithm six π₯ to the fourth power plus
one.

In our next example, weβll look at
how we can differentiate functions, which are a combination of both trigonometric
and logarithmic functions.

Differentiate π of π₯ equals five
times sin of five times the natural log of π₯.

This is a composite function. Itβs a function of a function. And we can, therefore, use the
chain rule to find its derivative. This says that if π¦ is some
function in π’ and π’ is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’
times dπ’ by dπ₯. Weβre going to let π’ be equal to
five times the natural log of π₯. And the derivative of the natural
log of π₯ is one over π₯. So dπ’ by dπ₯ is five times it;
itβs five times one over π₯, which is simply five over π₯. Instead of using π of π₯, letβs
use π¦. And this means that π¦ is equal to
five sin of π’. And the derivative of sin of π’ is
cos of π’. So dπ¦ by dπ’ is five cos π’. And according to the chain rule dπ¦
by dπ₯ is the product of these.

Referring back to the original
notation, we can say that π prime of π₯, the derivative, is five over π₯ times five
cos of π’. Weβll replace π’ with five times
the natural log of π₯. And we see that, in terms of π₯, π
prime of π₯ is five over π₯ times five cos of five times the natural log of π₯. And then we simplify and we see
theat π prime of π₯ is 25 over π₯ times the cosine of five times the natural log of
π₯.

In our final example, weβll
consider the higher order derivative of a logarithmic function.

Given that π of π₯ is equal to
nine times the natural log of π₯, find the fourth derivative of π of π₯.

This question is asking us to
differentiate our function with respect to π₯ four times. So letβs do this and see if we can
spot any patterns. Weβll begin by finding the first
derivative, π prime of π₯. We know that the derivative of the
natural log of π₯ is one over π₯, so π prime of π₯ is nine times one over π₯, which
is simply nine over π₯. We can write this alternatively as
nine π₯ to the power of negative one, which is going to allow us to find the next
derivative. Thatβs π double prime or π prime
prime of π₯. And he We recall the power rule for
differentiation. We multiply the expression by the
power and then subtract one from the power. And this means π double prime of
π₯ and the derivative of nine π₯ to the power of negative one is negative one times
nine π₯ to the power of negative two. Thatβs negative nine π₯ to the
negative two.

We can continue this to find the
third derivative, sometimes written as π prime prime prime or π with a three in
brackets here as shown. This is negative two times negative
nine π₯ to the power of negative three, which is positive 18π₯ to the negative
three. We repeat this process one more
time to find the fourth derivative. Thatβs negative three times 18π₯ to
the power of negative four. And the fourth derivative of nine
times the natural log of π₯ is negative 54π₯ to the power of negative four.

And this example demonstrates
general result. And that is the πth order
derivative of the natural logarithm function can be expressed as negative one to the
power of π minus one times π minus one factorial all over π₯ to the πth
power. This is a nice formula to spot,
though itβs also useful to be able to confidently apply the processes by
differentiating the natural logarithm function.

In this video, weβve learned that
the derivative of the natural logarithm of π₯ is equal to one over π₯ when π₯ is
greater than zero. Weβve seen that if we have the
natural logarithm of some function of π₯, we can use the chain rule to find the
derivative or, alternatively, we can quote the result that dπ¦ by dπ₯ is equal to π
prime of π₯ over π of π₯. We also saw that these rules can be
used in conjunction with the standard rules of differentiation, including the
product rule, the quotient rule, and the rules for finding higher order
derivatives.